Problem: $ F = \left[\begin{array}{rr}4 & -2 \\ 3 & 2 \\ 5 & 4\end{array}\right]$ $ A = \left[\begin{array}{rr}3 & 2 \\ 3 & -1\end{array}\right]$ What is $ F A$ ?
Solution: Because $ F$ has dimensions $(3\times2)$ and $ A$ has dimensions $(2\times2)$ , the answer matrix will have dimensions $(3\times2)$ $ F A = \left[\begin{array}{rr}{4} & {-2} \\ {3} & {2} \\ \color{gray}{5} & \color{gray}{4}\end{array}\right] \left[\begin{array}{rr}{3} & \color{#DF0030}{2} \\ {3} & \color{#DF0030}{-1}\end{array}\right] = \left[\begin{array}{rr}? & ? \\ ? & ? \\ ? & ?\end{array}\right] $ To find the element at any row $i$ , column $j$ of the answer matrix, multiply the elements in row $i$ of the first matrix, $ F$ , with the corresponding elements in column $j$ of the second matrix, $ A$ , and add the products together. So, to find the element at row 1, column 1 of the answer matrix, multiply the first element in ${\text{row }1}$ of $ F$ with the first element in ${\text{column }1}$ of $ A$ , then multiply the second element in ${\text{row }1}$ of $ F$ with the second element in ${\text{column }1}$ of $ A$ , and so on. Add the products together. $ \left[\begin{array}{rr}{4}\cdot{3}+{-2}\cdot{3} & ? \\ ? & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 2, column 1 of the answer matrix, multiply the elements in ${\text{row }2}$ of $ F$ with the corresponding elements in ${\text{column }1}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{3}+{-2}\cdot{3} & ? \\ {3}\cdot{3}+{2}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Likewise, to find the element at row 1, column 2 of the answer matrix, multiply the elements in ${\text{row }1}$ of $ F$ with the corresponding elements in $\color{#DF0030}{\text{column }2}$ of $ A$ and add the products together. $ \left[\begin{array}{rr}{4}\cdot{3}+{-2}\cdot{3} & {4}\cdot\color{#DF0030}{2}+{-2}\cdot\color{#DF0030}{-1} \\ {3}\cdot{3}+{2}\cdot{3} & ? \\ ? & ?\end{array}\right] $ Fill out the rest: $ \left[\begin{array}{rr}{4}\cdot{3}+{-2}\cdot{3} & {4}\cdot\color{#DF0030}{2}+{-2}\cdot\color{#DF0030}{-1} \\ {3}\cdot{3}+{2}\cdot{3} & {3}\cdot\color{#DF0030}{2}+{2}\cdot\color{#DF0030}{-1} \\ \color{gray}{5}\cdot{3}+\color{gray}{4}\cdot{3} & \color{gray}{5}\cdot\color{#DF0030}{2}+\color{gray}{4}\cdot\color{#DF0030}{-1}\end{array}\right] $ After simplifying, we end up with: $ \left[\begin{array}{rr}6 & 10 \\ 15 & 4 \\ 27 & 6\end{array}\right] $